log functions and exponential decay?

Question

The amount P of ozone in the atmosphere is currently decaying exponentially each year at a continuous rate of 0.27% (that is, k = -0.0027). How long will it take for half the ozone to disappear (that is, when will the amount be P / 2)? [Your answer is the half-life of ozone.]

f(x) = Pe^kx

Answer 1

Not too hard.
1. plug in what you know

P/2 = Pe^(-0.0027x)

2. cancel the Ps.
1/2 = e^(-0.0027x)

3. use some tricks with the ln and e to get x by itself
ln(1/2) = -0.0027x

4. solve for x
ln(1/2) / (-0.0027) = x

5. calculator, most definitely
x = 256.72


On step 3, try to remember this:
a = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.

Answer 2

y=yoe -0.0336t. Let OA = original amount and FA = final amount FA =OA (e^(-0.036t)) Divide both sides by OA e^(-0.036t) = (FA)/(OA) take the natural logarithm of both sides (the ln of e^x is x) -0.036t = ln(FA)/(OA) t = -ln((FA)/(OA))/0.036 .